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By using Descartes's sign rule , I can tell this polynomial $$x^{3}+3x-2\pi$$ has one real root.

But I want to know what that root is and what the factorization of it is. For the complex roots I can find them from the quadratic factor of this.

Any help on how to start thinking of the factorization of the polynomial are welcome.

Thanks.

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    $\begingroup$ See THIS for explicit roots of a cubic polynomial. $\endgroup$
    – Mark Viola
    Sep 4 '15 at 19:33
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    $\begingroup$ not possible algebraically? $\endgroup$
    – user118494
    Sep 4 '15 at 19:33
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    $\begingroup$ Yes! There is a closed form expression found HER. $\endgroup$
    – Mark Viola
    Sep 4 '15 at 19:35
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    $\begingroup$ An algebraic solution is messy but it does exist. You can in principle write a factorization, but it will be a horrible mess of a formula. It depends on your context, but my guess is that you would find the solution formula useless in this case. Do you really need an exact formula? $\endgroup$ Sep 4 '15 at 20:08
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    $\begingroup$ @user118494 You should research solving a depressed cubic. $\endgroup$
    – Lythia
    Sep 4 '15 at 20:13
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Things are really not as bad as people make them out to be. The first thing to recall is that polynomial equations up to and including degree $4$ can always be solved in terms of radicals. For the fully general case the formulas can get messy, but there are two simplifications:

  1. One can always assume that the second highest coefficient is zero by a Tschirnhaus-Transformation (linear substitution). This is already the case here and significantly simplifies the formulas in the cubic case.

  2. One can try to work with field theoretic methods directly. This should simplify things if the Galois group is sufficiently small compared to the fully symmetric group. This is not necessary here.

For this polynomial, it seems easiest to straightforwardly apply Cardano's formula.

(Cardano's formula) Let $K$ be a field with $\operatorname{char}(K) \ne 2,3$. Consider the cubic equation $x^3 + px + q =0$ where $p$, $q \in K$. The solutions are given by $x_1 = u + v$, $x_2 = \zeta^2 u + \zeta v$, $x_3 = \zeta u + \zeta^2 v$ where $\zeta$ is an arbitrary primitive cubic root of unity, and $$ \begin{align*} u &= \sqrt[3]{-q/2 + \sqrt{(p/3)^3 + (q/2)^2}} & v &=\sqrt[3]{-q/2 - \sqrt{(p/3)^3 + (q/2)^2}}, \end{align*} $$ where the third roots must be chosen subject to the sign condition $uv=-p/3$. The square root can be taken arbitrary, but it has to be the same one both times.

Note that a priori we would have three possibilities for each cubic root, giving a total of 9 possibilities. But the side condition on $uv$ reduces that to at most three distinct solutions.

Now, plugging in your equation we see $p=3$, $q=-2\pi$. (Incidentally, the coefficients are chosen in such a way that the formulas become particularly simply. We conclude that this was an exercise in a book or course.) Now choose $u = \sqrt[3]{\pi + \sqrt{1 + \pi^2}}$ and $v=\sqrt[3]{\pi - \sqrt{1 + \pi^2}}$ to be the real cubic roots. Then $u > 0$ and $v < 0$, so the sign condition is verified. A third root of unity is given by $\zeta = \frac{-1 + i \sqrt{3}}{2}$. The roots are now determined as above, in particular, $$ x_1=\sqrt[3]{\pi + \sqrt{1 + \pi^2}} + \sqrt[3]{\pi - \sqrt{1 + \pi^2}} $$ (with the cubic roots taken in $\mathbb R$) is the real root.

Note: Since $\pi$ is in fact transcendental over $\mathbb Q$, this is the same as factoring $g=x^3 + 3x -2y$ in the splitting field of $g$ over the rational function field $\mathbb Q(y)$.

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